

A069283


a(n) = 1 + number of odd divisors of n.


8



0, 0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 3, 0, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 3, 1, 0, 3, 1, 3, 2, 1, 1, 3, 1, 1, 3, 1, 1, 5, 1, 1, 1, 2, 2, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 1, 5, 0, 3, 3, 1, 1, 3, 3, 1, 2, 1, 1, 5, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 3, 1, 3, 1, 1, 2, 5, 2
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OFFSET

0,10


COMMENTS

Number of nontrivial ways to write n as sum of at least 2 consecutive integers. That is, we are not counting the trivial solution n=n. E.g., a(9)=2 because 9 = 4 + 5 and 9 = 2 + 3 + 4. a(8)=0 because there are no integers m and k such that m + (m+1) + ... + (m+k1) = 8 apart from k=1, m=8.  Alfred Heiligenbrunner (alfred.heiligenbrunner(AT)gmx.at), Jun 07 2004
Also number of sums of sequences of consecutive positive integers excluding sequences of length 1 (e.g., 9 = 2+3+4 or 4+5 so a(9)=2). (Useful for cribbage players.)  Michael Gilleland, Dec 29 2002
Let M be any positive integer. Then a(n) = number of proper divisors of M^n + 1 of the form M^k + 1.
This sequence gives the distinct differences of triangular numbers Ti giving n : n = Ti  Tj; none if n = 2^k. If factor a = n or a > (n/a  1)/2 : i = n/a + (a  1)/2; j = n/a  (a+1)/2. Else : i = n/2a + (2a  1)/2; j = n/2a  (2a  1)/2. Examples: 7 is prime; 7 = T4  T2 = (1 + 2 + 3 + 4)  (1 + 2) (a = 7; n/a = 1). The odd factors of 35 are 35, 7 and 5; 35 = T18  T16 (a = 35) = T8  T1 (a = 7) = T5  T7 (a = 5). 144 = T20  T11 (a = 9) = T49  T46 (a = 3).  M. Dauchez (mdzzdm(AT)yahoo.fr), Oct 31 2005
Also number of partitions of n into the form 1 + 2 + ...( k  1) + k + k + ... + k for some k >= 2. Example: a(9) = 2 because we have [2, 2, 2, 2, 1] and [3, 3, 2, 1].  Emeric Deutsch, Mar 04 2006
a(n) is the number of nontrivial runsum representations of n, and is also known as the politeness of n.  Ant King, Nov 20 2010


REFERENCES

Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed. (AddisonWesley, 1994), see exercise 2.30 on p. 65.
Apostol, Tom M.; Sums of Consecutive Positive Integers, The Mathematical Gazette, Vol. 87, No. 508, (March 2003), pp. 98101.


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
A. Heiligenbrunner, Sum of adjacent numbers (in German).
Wikipedia, Polite Number


FORMULA

a(n) = 0 if and only if n = 2^k.
a(n) = 1 if and only if n = 2^k x p where k >= 0 and p is an odd prime.  Ant King, Nov 20 2010
G.f.: sum(k>=2, x^(k(k + 1)/2)/(1  x^k) ).  Emeric Deutsch, Mar 04 2006
If n = 2^k p1^b1 p2^b2 ... pr^br, then a(n) = (1 + b1)(1 + b2) ... (1 + br)  1.  Ant King, Nov 20 2010


EXAMPLE

a(14) = 1 because the divisors of 14 are 1, 2, 7, 14, and of these, two are odd, 1 and 7, and 1 + 2 = 1.
a(15) = 3 because the divisors of 15 are 1, 3, 5, 15, and of these, all four are odd, and 1 + 4 = 3.
a(16) = 0 because 16 has only one odd divisor, and 1 + 1 = 0.
Using Ant King's formula: a(90) = 5 as 90 = 2^1 * 3^2 * 5^1, so a(90) = (1 + 2) * (1 + 1)  1 = 5.  Giovanni Ciriani, Jan 12 2013
x^3 + x^5 + x^6 + x^7 + 2*x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + ...


MAPLE

g:=sum(x^(k*(k+1)/2)/(1x^k), k=2..20): gser:=series(g, x=0, 115): seq(coeff(gser, x, n), n=0..100); # Emeric Deutsch, Mar 04 2006


MATHEMATICA

g[n_] := Module[{dL = Divisors[2n], dP}, dP = Transpose[{dL, 2n/dL}]; Select[dP, ((1 < #[[1]] < #[[2]]) && (Mod[ #[[1]]  #[[2]], 2] == 1)) &] ]; Table[Length[g[n]], {n, 1, 100}]
Table[Length[Select[Divisors[k], OddQ[#] &]]  1, {k, 100}] (* Ant King, Nov 20 2010 *)


PROG

(Haskell)
a069283 0 = 0
a069283 n = length $ tail $ a182469_row n
 Reinhard Zumkeller, May 01 2012
{a(n) = if( n<1, 0, sumdiv( n, d, d%2)  1)} /* Michael Somos, Aug 07 2013 */


CROSSREFS

Cf. A001227, A062397, A057934. Equals A001227(n)  1. A138591.
Cf. A182469.
Sequence in context: A046214 A232088 A115413 * A033630 A220122 A101446
Adjacent sequences: A069280 A069281 A069282 * A069284 A069285 A069286


KEYWORD

nonn


AUTHOR

Reinhard Zumkeller, Mar 13 2002


EXTENSIONS

Edited by Vladeta Jovovic, Mar 25 2002


STATUS

approved



